Timeline for Construct a PDE solution from a net of approximations
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 26, 2015 at 6:54 | comment | added | David Ketcheson | Sorry, didn't read carefully. | |
| Aug 25, 2015 at 18:52 | comment | added | Alex M. | @DavidKetcheson: Since I already work with distributions, what would "weak" mean? Can one get "weaker" than distributions? | |
| Aug 23, 2015 at 8:41 | comment | added | David Ketcheson | I guess you are only interested in strong solutions? There are certainly examples of sequences of continuous functions that satisfy your conditions, but their limit is discontinuous, so not a strong solution of the PDE. | |
| Aug 22, 2015 at 12:33 | comment | added | Dirk | @AlexM. I guess that your limit is something like "largest set in the partition goes to zero". Why not pick sequences of such converging partitions and show that limits are independent of the subsequences? | |
| Aug 22, 2015 at 12:30 | comment | added | Dirk | @IgorKhavkine This counterexample does not work with unique solutions (as requested by the OP or the edited question with explicit boundary conditions). | |
| Aug 22, 2015 at 12:29 | comment | added | Alex M. | @IgorKhavkine: To put my problem in your notations, I have $f(x_i) \to 0$ but I do not have $x$. I have good reasons to believe that it must exist, though. Concerning your counterexample, I have added further data concerning the boundary condition that rules it out - thank you for making the statement more precise. | |
| Aug 22, 2015 at 12:27 | comment | added | Alex M. | @Dirk: In my problem, the index set $I$ is the set of all the partitions of a given set, hence nets instead of sequences. | |
| Aug 22, 2015 at 12:25 | history | edited | Alex M. | CC BY-SA 3.0 |
Made the statement more specific
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| Aug 22, 2015 at 11:11 | comment | added | Igor Khavkine | @Dirk, if that is the case, then there are obvious counter examples like $P=d/dx$ and $u_i = i$ (constant functions) with $i \in \mathbb{N}$. | |
| Aug 22, 2015 at 11:02 | comment | added | Dirk | @IgorKhavkine I guess the difference is that the OP does not assume that $x_i$ converges to $x$ but only that $f(x)$ converges to zero and wants to conclude the convergence of the $x_i$. | |
| Aug 22, 2015 at 11:00 | comment | added | Dirk | I can't imagine a situation in which one naturally has a net of a solution of a PDE which is not a sequence. But maybe that's only my limited imagination… | |
| Aug 22, 2015 at 10:58 | comment | added | Igor Khavkine | How is this different from asking whether $f(x_i)$ converges to $f(x)$ when $x_i$ converges to $x$ on a general topological space? Obviously they agree when $f$ is continuous. | |
| Aug 22, 2015 at 9:01 | history | asked | Alex M. | CC BY-SA 3.0 |